Question: $f(x)=x^2-10x-56$ 1) What are the zeros of the function? Write the smaller $x$ first, and the larger $x$ second. $\text{smaller }x=$
Answer: To find the zeros of the function, we need to solve the equation $f(x)=0$. We can do that by factoring $f(x)$. $\begin{aligned} x^2-10x-56&=0 \\\\ (x-14)(x+4)&=0 \\\\ x-14=0&\text{ or }x+4=0 \\\\ x={14}&\text{ or }x={-4} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $x$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }x\text{-coordinate}&=\dfrac{({14})+({-4})}{2} \\\\ &={5} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $f({5})$ : $\begin{aligned} f({5})&=({5})^2-10({5})-56 \\\\ &=25-50-56 \\\\ &=-81 \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }x&=-4 \\\\ \text{larger }x&=14 \end{aligned}$ The vertex of the parabola is at $(5,-81)$